/*
Let's call S the (infinite) string that is made by concatenating the consecutive positive integers (starting from 1)  written down in base 10. 
Thus, S = 1234567891011121314151617181920212223242...


It's easy to see that any number will show up an infinite number of times in S.


Let's call f(n) the starting position of the nth occurrence of n in S. 
For example, f(1)=1, f(5)=81, f(12)=271 and f(7780)=111111365.


Find ∑f(3k) for 1≤k≤13.

Anser:
Time:
*/
package main

import (
	"fmt"
	"time"
)

func main() {
	tstart := time.Now()



	tend := time.Now()
	fmt.Println(tend.Sub(tstart))
}